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Prove that for positive values of $a,b,c$: if $a^2,b^2,c^2$ are the sides of a triangle, then there exists a triangle with sides $a,b,c$. (Hint: You may find it helpful to recall that three segments can only form a triangle if the sum of the lengths of any two of them is larger than the length of the remaining segment.)

Let's try to argue by contradiction...

Assume that there does not exist a triangle with sides $a,b,c$.

Recalling that three segments can only form a triangle if the sum of the lengths of any two of them is larger than the length of the third, it must be the case that the sum of $a$ and $b$ must be less than or equal to $c$. Consequently, $$
\begin{align}
a+b & \le c\\
(a+b)^2 & \le c^2\\
a^2 + 2ab + b^2 & \le c^2
\end{align}$$
Now, note that since $a$ and $b$ are positive, $2ab$ must also be positive. Thus, we can subtract $2ab$ from just the left side and still preserve the direction of the inequality, yielding $$a^2 + b^2 \lt c^2$$ Hence, there does not exist a triangle with sides $a^2,b^2,c^2$. But this contradicts the given fact that there does exist a triangle with these side lengths. Consequently, our assumption must be rejected, and the opposite must be true:

There exists a triangle with sides $a,b,c$.

QED.

Note, with some slight modification, we can also argue the result directly...

If there is a triangle with sides $a^2$, $b^2$, and $c^2$, it must be the case that

$$a^2 + b^2 \gt c^2$$

In considering a triangle of sides $a$, $b$, and $c$, we first note that $a$ and $b$ must be positive as they represent lengths. Consequently, $2ab > 0$ and we can add this value to the left side above, while preserving the inequality. Hence,

In a similar way, we can argue that it must be the case that $a^2 + c^2 \gt b^2$, which leads to $a+c \gt b$; and $b^2 + c^2 \gt a^2$, which leads to $b+c>a$. Since the sum of any two of the values $a$, $b$, and $c$ sum to something greater than the third, there does indeed exist a triangle with these side lengths.